# Question 9.5: Steel billets of 35 mm diameter and 500 mm length are to be ......

Steel billets of 35 mm diameter and 500 mm length are to be heated to 1200°C by direct resistance heating.

Heating time (sec)      30

Density (g/cm³)      7.6

Mean thermal conductivity (λ.w/m °C)      35.7

Mean specific heat (kJ/kg °C)      0.576

Mean resistivity (σ ⋅ ohm ⋅ cm)      61 × $10^{-6}$

Ambient temperature (°C)      30

Magnetic permeability ($μ_r$)      100

Permeability of air ($μ_o$)     4$\pi$ × $10^{-7}$

Now determine:
1. Power required to heat the billet from 30 to 1200°C.
2. Power lost by radiation and total power required.
3. Current required from a 100 V DC or AC supply.
4. Effective resistance for DC and AC heating.

Step-by-Step
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Volume of billet

= $=\frac{\pi \times 3.5^2 \times 50}{4}=480 \ \mathrm{~cm}^3$
Surface area (curved surface)
= π × 3.5 × 50
= 550 cm²
= 0.055 m²
Weight
= 480 × 7.6
= 3648 gm
= 3.648 kg
Resistance

\begin{aligned} \sigma \frac{\ell}{A} & =\frac{61 \times 10^{-6} \times 50}{38.48} \\ & =80 \times 10^{-6} \ \Omega \end{aligned}
Heat required to raise the temperature from 30 to 1200°C = mass × sp. heat × temp. difference
= 3.648 × 10³ × 0.576 × (1200 − 30)
= 2.458 × 106 J
If heating is done in 30 sec

Power = $\frac{2.458 \times 10^6}{30}$

= 85 kW

Assume a mean surface temperature 800°C and an emissivity 0.8

\begin{aligned}Q & =5.67 \times 0.8 \times\left[\left(\frac{1200+273}{100}\right)^4-\left(\frac{30+273}{100}\right)^4\right] \times 0.055 \\& =1.1785 \times 10^4 \ \mathrm{~W} \\& \simeq 12 \ \mathrm{~kW}\end{aligned}

Total power
= 85 + 12
= 97 kW~100 kW

Note : There may be additional losses in grips.

If a 100 V DC power supply is available

Current required = $\frac{100 \times 10^3}{1000}$

= 1000 A

The radiation loss is very high. It can be reduced by keeping the billet in an enclosure or heating for a shorter duration.
Note that the calculated radiation loss is at 1200°C surface temperature which is attained at the end of the heating time.
If the power available is AC at 50 Hz, the effective resistance will change from $R_{DC}$ to $R_{AC}$.
$R_{DC}$ = 80 × $10^{-6}$ Ω (as calculated earlier)
$\rho$ ⋅ (30–760°C) = 57× $10^{-6}$ ohm.cm
$\rho$ ⋅ (760–1200°C) = 120 ×$10^{-6}$ ohm.cm
$\mu_o=4 \pi \times 10^{-7} \quad \mu_r=100$
We can get $R_{AC}$ by using the formula
$R_{AC} = \rho \frac{\ell}{A}$

where a is the area of annulus through which most of the current flows. This will be circular ring of outer diameter 3.5 cm and inner diameter = 3.5 − 2δ, where δ is the depth of penetration.
For the temperature range 30–760°C

$\delta=5030 \sqrt{\frac{\rho}{\mu r \cdot f}} \\ \delta=5030 \sqrt{\frac{57 \times 10^{-6}}{100 \times 50}}$

= 0.53 cm or 5.3 mm

Inner diameter = $3.5-2(2 \times \delta)$

= 3.5 – 2 (2× 0.53)

= 1.38 cm

Area = $\frac{\pi\left(3.5^2-1.38^2\right)}{4}$

= 8.125 cm²

\begin{aligned}R_{A C} & =\frac{57 \times 10^{-6} \times 50}{8.125} \\& =350 \times 10^{-6}\end{aligned}

or    = 35 × $10^{-5}$ ohm (30-760°C).

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